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NCERT Solutions class 12 Maths Exercise 4.3 Ch 4 Determinants

NCERT Solutions for Class 12 Maths Exercise 4.3 Chapter 4 Determinants – FREE PDF Download

NCERT Solutions for Class 12 Maths Chapter 4 – Determinants is a sure-shot way of obtaining the complete marks in the particular chapter for Board Exam 2019- 2020. CoolGyan provides you with Free PDF download of the same solved by Expert Teachers as per NCERT (CBSE) Book guidelines. They provide the students with precise and to the point answers which fetch very good marks in Board Exams. Download today the NCERT CBSE Solutions for Class 12 Maths Chapter 4 – Determinants to achieve your goal score. Class 12 Maths Chapter 4 – Determinants solved by Expert Teachers as per NCERT (CBSE) Book guidelines.

NCERT Solutions for Class 12 Maths Chapter 4 – Determinants



1. Find the area of the triangle with vertices at the points given in each of the following:

(i) (1, 0), (6, 0), (4, 3)

(ii) (2, 7), (1, 1), (10, 8)

(iii) 

Ans. (i) Area of triangle = Modulus of  

 =  sq. units

(ii) Area of triangle = Modulus of 

 =  sq. units

(iii) Area of triangle = Modulus of 

 = 15 sq. units


2. Show that the points A B C are collinear.

 

Ans. Area of triangle ABC = Modulus of  =  

 = 0

Therefore, points A, B and C are collinear.


3. Find values of  if area of triangle is 4 sq. units and vertices are:

 

(i) 

(ii) 

Ans. (i) Given: Area of triangle = Modulus of  = 4 

 Modulus of  = 4

 

 

 

 

Taking positive sign, 

 

Taking negative sign, 

 

(ii) Given: Area of triangle = Modulus of  = 4

  = 4

 

 

 

 

Taking positive sign, 

 

Taking negative sign, 

 


4. (i) Find the equation of the line joining (1, 2) and (3, 6) using determinants.

 

(ii) Find the equation of the line joining (3, 1) and (9, 3) using determinants.

Ans. (i) Let P be any point on the line joining the points (1, 2) and (3, 6). 

Then, Area of triangle that could be formed by these points is zero.

 Area of triangle = Modulus of  = 0

 Modulus of  = 0

 

 

 

  which is required line.

(ii) Let P be any point on the line joining the points (3, 1) and (9, 3).

Then, Area of triangle that could be formed by these points is zero.

 Area of triangle = Modulus of  = 0

 Modulus of  = 0

 

 

 

   which is required line.


5. If area of triangle is 35 sq. units with vertices  and  Then  is:

 

(A) 12

(B) 

(C) 

(D) 

Ans. Given: Area of triangle = Modulus of  = 35 

 Modulus of  = 35

 

 

 

 

 

Taking positive sign, 

 

Taking negative sign, 

 

Therefore, option (D) is correct.